How to generate normally distributed random from an integer range?

Question

Given the start and the end of an integer range, how do I calculate a normally distributed random integer between this range?

I realize that the normal distribution goes into -+ infinity. I guess the tails can be cutoff, so when a random gets computed outside the range, recompute. This elevates the probability of integers in the range, but as long as the this effect is tolerable (<5%), it's fine.

public class Gaussian
{
    private static bool uselast = true;
    private static double next_gaussian = 0.0;
    private static Random random = new Random();

    public static double BoxMuller()
    {
        if (uselast) 
        { 
            uselast = false;
            return next_gaussian;
        }
        else
        {
            double v1, v2, s;
            do
            {
                v1 = 2.0 * random.NextDouble() - 1.0;
                v2 = 2.0 * random.NextDouble() - 1.0;
                s = v1 * v1 + v2 * v2;
            } while (s >= 1.0 || s == 0);

            s = System.Math.Sqrt((-2.0 * System.Math.Log(s)) / s);

            next_gaussian = v2 * s;
            uselast = true;
            return v1 * s;
        }
    }

    public static double BoxMuller(double mean, double standard_deviation)
    {
        return mean + BoxMuller() * standard_deviation;
    }

    public static int Next(int min, int max)
    {
        return (int)BoxMuller(min + (max - min) / 2.0, 1.0); 
    }
}

I probably need to scale the standard deviation some how relative to the range, but don't understand how.

Answer:

    // Will approximitely give a random gaussian integer between min and max so that min and max are at
    // 3.5 deviations from the mean (half-way of min and max).
    public static int Next(int min, int max)
    {
        double deviations = 3.5;
        int r;
        while ((r = (int)BoxMuller(min + (max - min) / 2.0, (max - min) / 2.0 / deviations)) > max || r < min)
        {
        }

        return r;
    }

Answer 1

If the Box-Muller method returns a "standard" normal distribution, it will have mean 0 and standard deviation 1. To transform a standard normal distribution, you multiply your random number by X to get standard deviation X, and you add Y to obtain mean Y, if memory serves me correctly.

See the Wikipedia article's section on normalizing standard normal variables (property 1) for a more formal proof.

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In response to your comment, the rule of thumb is that 99.7% of a normal distribution will be within +/- 3 times the standard deviation. If you need a normal distribution from 0 to 100 for instance, than your mean will be halfway, and your SD will be (100/2)/3 = 16.667. So whatever values you get out of your Box-Muller algorithm, multiply by 16.667 to "stretch" the distribution out, then add 50 to "center" it.

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John, in response to your newest comment, I'm really not sure what is the point of the Next function. It always uses a standard deviation of 1 and a mean of halfway between your min and max.

If you want a mean of Y, with ~99.7% of the numbers in the range -X to +X, then you just call BoxMuller(Y, X/3).

Answer 2

Well, the -2*sigma..+2*sigma will give you 95% of the bell curve. (check the "Standard deviation and confidence intervals" section in the already mentioned wiki article).

So modify this piece:

return (int)BoxMuller(min + (max - min) / 2.0, 1.0);

and change 1.0 (standard deviation) to 2.0 (or even more if you want more than 95% coverage)

return (int)BoxMuller(min + (max - min) / 2.0, 2.0);